An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000Aᵒ. What is the maximum kinetic energy of the emitted photoelectron?

Solution

Physics sir · 4 steps

The photon carries the electron's 3 eV of kinetic energy plus the binding energy it loses on capture.

1. Energy before capture. Far from the proton the potential energy is zero, so the electron's total energy is just its kinetic energy, +3 eV.
2. Energy after capture. The second excited state is n = 3, because the ground state is n = 1. Its energy is E₃ = -13.6/3² = -1.51 eV.
3. Photon energy. The photon takes away the energy difference: Eₚₕₒₜₒₙ = 3 - (-1.51) = 4.51 eV.

   Kₘₐₓ = Eₚₕₒₜₒₙ - φ = 4.51 - 12400/4000 = 4.51 - 3.1 = 1.41 eV

4. So. The work function from the 4000 Å threshold is 3.1 eV, and the photon has more energy than that. The maximum kinetic energy is 1.41 eV, option (b).

The idea to keep. In a capture, the photon energy is the starting energy minus the final energy. Count the starting kinetic energy, and remember that the second excited state means n = 3.

Answer from the NTA final answer key. Solution by PrepWiser.